# If X~N(μ,σ^2) then what's the distribution for exp(X) ?

1. Mar 1, 2013

### stevenytc

If X is a r.v. that follows that Gaussian distribution with mean μ and variance σ^2, how do I find the distribution function for exp(X) where exp() is the exponent function.

2. Mar 1, 2013

### jbunniii

If $Y = \exp(X)$, then the cdf of $Y$ is $F_Y(\alpha) = P(\exp(X) \leq \alpha)$. As $\log$ is strictly monotonically increasing, this is equivalent to $P(X \leq \log(\alpha))$, which is $F_X(\log(\alpha))$, the cdf of $X$ evaluated at $\log(\alpha)$.

Then you can find the pdf of $Y$ by differentiating $F_Y(\alpha) = F_X(\log\alpha))$ to obtain $f_Y(\alpha) = f_X(\log(\alpha)) / \alpha$. This is valid for all positive $\alpha$; if $\alpha$ is negative then the pdf is zero since $\exp$ is never negative.

Proceeding from here should be straightforward as there is a closed form expression for $f_X$.

3. Mar 1, 2013

### awkward

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